question_answer

The perimeter of a right triangle is 12 cm. The hypotenuse is 5 cm. The other two sides and area of the triangle are

A) 3, 4 and \[6\,\,\text{c}{{\text{m}}^{\text{2}}}\]

B)  4, 3 and \[12\,\,\text{c}{{\text{m}}^{\text{2}}}\]

C)  6, 2 and \[6\,\,\text{c}{{\text{m}}^{\text{2}}}\]

D)  None of these

Correct Answer: A

Solution :

Perimeter = 12 cm

\[\therefore \]      \[a+b+5=12\]

\[\Rightarrow \]   \[a+b=7\,\,\text{cm}\]

Also by Pythagoras theorem,

\[{{a}^{2}}+{{b}^{2}}=25\]

Also, \[{{(a+b)}^{2}}+{{(a-b)}^{2}}=2\,\,({{a}^{2}}+{{b}^{2}})\]

\[{{(a-b)}^{2}}=2\,\,({{a}^{2}}+{{b}^{2}})-{{(a+b)}^{2}}\]

\[=2\,\,(25)-{{(7)}^{2}}=50-49=1\]

\[\Rightarrow \]   \[a-b=\pm \,\,1\]

\[\therefore \]      \[a+b=7\]and \[a-b=1\]or \[a-b=-\,\,1\]

\[\Rightarrow \]   \[a=4\,\,\text{cm}\]and \[a=3\,\,\text{cm}\]

\[\therefore \]Area of triangle \[=\frac{1}{2}\times 3\times 4=6\,\,\text{c}{{\text{m}}^{2}}\]